Question

A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. What is the empirical formula of the phosphorus selenide?

Answer 1

Answer:

Explanation:

Phosphorus:

mass = 45.2 mg

molar mass = 30.97 mg / mmol

moles = 45.2 mg / (30.97 mg / mmol) = 1.46 mmol ≈ 1.5

 

Selenium:

mass = 131.6 mg – 45.2 mg = 86.4 mg

molar mass = 78.96 mg / mmol

moles = 86.4 mg / (78.96 mg / mmol) = 1.10 mmol ≈ 1

 

P1.5Se1

To get the empirical formula, we multiply with 2 to get whole numbers:

P3Se2